Question 367018
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What you are doing wrong is guessing.  You lucked out and guessed the outer dimensions from which you could have simply subtracted 4 feet (2 foot wide flower bed all around) to get the slab dimensions.  But in general this would have been a poor strategy for this sort of problem if you had been given dimensions that did not come out to nice round numbers.


Let's try a systematic approach:


Let *[tex \Large x] represent the width of the slab.  Then *[tex \Large x\ +\ 3] must represent the length of the slab.  Since the flower bed is two feet wide and hence adds two feet to EACH side and EACH end of the overall area.  That means that the width dimension of the outside edge of the flower bed is *[tex \Large x\ +\ 4] and the length dimension must be *[tex \Large x + 7].


Using *[tex \Large A\ =\ lw], and knowing that the overall area is 270, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 4)(x\ +\ 7)\ =\ 270]


A little algebra gets us to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 11x\ -\ 242\ =\ 0]


Solve the quadratic for *[tex \Large x].  Discard the negative root since we are looking for a positive measure of length.  The positive root is the width of the concrete slab and 3 more is the length.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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